# Thread: Integration: Trigonometric substitution

1. ## Integration: Trigonometric substitution

Hey

The problem i am stuck on is: find the integral of (x^2)/(sqrt(9-x^2))

I have managed to complete the problem to the second to last step where i get (9 theta)/2-9/4 sin(2 theta),

since i substituted at the beginning x=3sin(theta) i can rearrange for theta and get the first part = 9/2sin^-1(x/3).
What I am struggling with is to complete the 9/4sin(2theta) by putting this back in terms of x.
The answer i am looking for is 1/2 x sqrt(9-x^2) so if someone could explain how to get 9/4 sin(2 theta) to 1/2 x sqrt(9-x^2) that be be extremely helpful.

Thanks

2. Originally Posted by Sam1111
Hey

The problem i am stuck on is: find the integral of (x^2)/(sqrt(9-x^2))

I have managed to complete the problem to the second to last step where i get (9 theta)/2-9/4 sin(2 theta),

since i substituted at the beginning x=3sin(theta) i can rearrange for theta and get the first part = 9/2sin^-1(x/3).
What I am struggling with is to complete the 9/4sin(2theta) by putting this back in terms of x.
The answer i am looking for is 1/2 x sqrt(9-x^2) so if someone could explain how to get 9/4 sin(2 theta) to 1/2 x sqrt(9-x^2) that be be extremely helpful.

Thanks
$\tfrac{9}{4}\sin\!\left(2\theta\right)=\tfrac{9}{4 }\left(2\sin\theta\cos\theta\right)=\tfrac{9}{2}\s in\theta\cos\theta$. From your substitution, we know that $x=3\sin\theta\implies \sin\theta=\frac{x}{3}$. With this, it is easily shown that $\cos\theta=\frac{\sqrt{9-x^2}}{3}$.

Thus, $\tfrac{9}{2}\sin\theta\cos\theta=\tfrac{9}{2}\left (\tfrac{1}{3}x\cdot\tfrac{1}{3}\sqrt{9-x^2}\right)=\tfrac{1}{2}x\sqrt{9-x^2}$

Does this make sense?